CHAPTER ONECHAPTER ONE

SEMICONDUCTORSSEMICONDUCTORS

Copyright, 2006 © Ahmed S. Bouazzi

2e A G.I. Module Energie Solaire

المدرسة الوطنية

للمهندسين بتونس

The Crystal Lattice of The Crystal Lattice of SiliconSilicon

Each silicon atom is situated at the center of a tetrahedron and connected to four other atoms occupying the summit of the tetrahedron.

A two dimension A two dimension representation of the silicon representation of the silicon

crystal structurecrystal structure

Intrinsic silicon

Si Si

SiSi

Si Si

Si

Si

Si Si

Si Si

SiSi

Si

Si

Each silicon atom is situated at the center of four other atoms.

N-type SemiconductorsN-type Semiconductors

n-doped silicon

P

P

P

P Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

SiSi

Si

Si

Si

P-type SemiconductorsP-type Semiconductors

p-doped silicon

Si Si

SiSi

Si Si

Si

Si

Si Si

Si Si

Si

B

B

B

Energy

Interatomic distance

Eg

Permitted levels

(a)

(b)

The GapThe Gap

Fermi LevelFermi Level

1exp

1)(

kTEE

EFF

[ F(E) is the probability for an electron to be in the E energy level]

In intrinsic silicon, EF is situated in the middle of the gap. In doped silicon, the Fermi level goes up or down depending on the electron concentration.

Semiconductor DopingSemiconductor Doping

E ia

E id E

F

E c

E g

E v

E F

n doped silicon p doped silicon

The doping atoms create localized levels inside the band gap.

Energy

vEEF

cE

cE

vE

SemiconductorMetal

Electron-hole PairsElectron-hole Pairs

Intrinsic Carrier Intrinsic Carrier ConcentrationConcentration

= 42in

3

2

2

h

kT

kT

Eg

for silicon at 300 K, ni2 = 2x1020 cm-6

np =2in

(memh)3/2 exp

• In n-type silicon:

Minority and Majority Minority and Majority CarriersCarriers

n = ND

pno ND = ni2

pno = ni2/ ND

• In p-type silicon:

p = NA

npo NA = ni2

npo = ni2/ NA

• In n-type silicon:

Minority and Majority Minority and Majority Carriers in Excess Carriers in Excess

nn0 = ND ≈ 1016 – 1018 cm-3

pn0 ND = ni2; pno = ni

2/ ND ≈ 2×104 – 2×102 cm-3

Creating n = p (≈ 1011 – 1014 cm-3) electron-hole pairs will give:

nn = nn0 + n ≈ ND

and pn = pn0 + p ≈ p

Minority and Majority Carriers in Minority and Majority Carriers in Excess Excess

• In p-type silicon:

pp0 = NA ≈ 1016 – 1018 cm-3

np0 NA = ni2; npo = ni

2/ NA ≈ 2×104 – 2×102 cm-3

Creating n = p (≈ 1011 – 1014 cm-3) electron-hole pairs will give:

pp = pp0 + p ≈ NA

and np = np0 + n ≈ n

Lifetime and Lifetime and RecombinationRecombination

N = Noexp

t•In the bulk:

•At the surface:

Jsur = q(np - npo)S

is the lifetime of the minority carriers.

S is the surface recombination velocity of minority carriers.

Diffusion LengthDiffusion Length

L = D

•The diffusion length is the free path of the minority carriers before recombination.

is the lifetime of the minority carriers.

D is the diffusion constant of minority carriers.

Absorption CoefficientAbsorption Coefficient

xo e 1

The absorbed quantity of photonsat the depth x is:

0 is the flux of photons arriving at the surface of the semiconductor and x is the depth.

x

0

0

xoe

Drift Minority Carriers Drift Minority Carriers Current in a Current in a

SemiconductorSemiconductor

• Electrons:

Jn = qnpµnE and n = qnpµn

Drift Minority Carriers Drift Minority Carriers Current in a Current in a

SemiconductorSemiconductor

•Holes:

Jp = qpnµpE and p = qpnµp

Jn = qDnnp(x,y,z)

Jp = – qDppn(x,y,z)

Diffusion Minority Carriers Diffusion Minority Carriers Current in a Current in a

SemiconductorSemiconductor

J = Jn + Jp

• Electrons:

Jn = qnpµnE + qDnnp(x,y,z)

• Holes:

Jp = qpnµpE – qDppn(x,y,z)

Total Minority Carriers Total Minority Carriers CurrentCurrent

Ec

Ev

EF

Eg p

n

Junction plane

p-n Junctionp-n Junction

Depletion region

PHOTOVOLTAIC EFFFECT PHOTOVOLTAIC EFFFECT (1)(1)

The p-n Junction

E c

E v

E F

p

n

Electrons Light HHHooollleeesss

E c

E v

E f

Electrons current

Holes current

The photocurrent under illumination

PHOTOVOLTAIC EFFFECT PHOTOVOLTAIC EFFFECT (2)(2)

PHOTOVOLTAIC EFFFECT PHOTOVOLTAIC EFFFECT (3)(3)

The photovoltage under illumination

E c

E v

E F

p

n The

photovoltage

2

1

2

1

112

DA

B

NNq

Vw

Depletion RegionDepletion Region

Where the built-in voltage is defined by:

2Log

i

DAB n

NN

q

kTV

E c

E v

EF

p

nEc

Ev

EF

p

n

Polarization of a p-n Polarization of a p-n JunctionJunction

direct inverse

J (A/cm2)

U (V)

0.5 1.0

1exp

kT

qUJJ o

I-V CharacteristicI-V Characteristic

n

np

p

pno L

Dn

L

DpqJ

n

np

p

pno L

Dn

L

DpqJ

Saturation CurrentSaturation Current

np = ; pn =A

i

N

n2

D

i

N

n2

Dp

p

An

ni NL

D

NL

Dqn2

; Lk= kkD

Jo =

Metal/semiconductor junction

Ec

Ev

SemiconductorEF

metal

Ec

EF

Ev

(a)

(b)

(c)

Schottky Diode and Ohmic Schottky Diode and Ohmic ContactsContacts